891 2 11 review help

What are cohomology rings good for?

I am studying some concepts of algebraic topology myself, and I read lately a bit about cohomology rings (created by the direct sum ofcohomology groups) but besides all definitions I could not find any information.

What are they good for?How do we use them?Could someone give me a very short descriptive example?

Thank you

(algebraic-topology) (cohomology)

edited May 17 '11 at 19:18Raeder1,185 1 11 22

asked May 17 '11 at 19:05Jacek31 1

The example on the wiki page for cup product will be helpful: en.wikipedia.org/wiki/Cup_product Robert Bell

May 17 '11 at 19:38

2

to add to the excellent answers below - a couple of classic cases where the (co)homology and fundamental groupscoincide, yet the spaces are not isomorphic: (1) and , and (2) and . (2) S2 S1 S1 S1 S1 S1 S2 S3 S1 S2is a bit harder to show (at least it was for me yesterday) Juan S May 18 '11 at 4:00

3 Answers

The main idea behind the cohomology ring is that you have an extra structure that allows youto say more about your space. In some instances, you can use it to say that two spaces whichhave isomorphic (co)homology groups are different because they have different cohomologyrings, and in some instances you can infer information about the cohomology groups byknowing that there is an additional structure.

For an example of the first case, the cohomology of real projective space is. This distinguishes projective space from a wedge product

, because the multiplication in the corresponding cohomology ring is trivial.( ;/2) /2[x]/( )H Pn xn+1

in Si

For an example of the second case, suppose that a space has a product on it satisfying somemild conditions, making it an . Then the cohomology ring will be a Hopf algebra,. If thespace is also a connected, compact manifold, then (taking cohomology over a field ofcharacteristic zero) one can use the classification of connective, finite type, commutative hopfalgebras (the Milnor-Moore theorem) to say what the ring looks like. In particular, it must bethe alternating algebra on odd-degree generators. This is enough to show that there are noodd-dimensional analogues of the quaternions.

H-space

I am blanking on truly simple examples which exemplify the power of cohomology rings, but itshould at least be plausible that, just as you can say useful things by knowing that a set has agroup structure, you can say useful things by knowing that a graded vector space has a graded-commutative ring structure.

One other application worth mentioning is that the cohomology ring (plus Poincare duality)allows us to study intersections in an algebraic manner.

Suppose that is an dimensional compact oriented manifold, and that and are orientedsubmanifolds with . If we perturb and slightly (so that the intersectionis , they will intersect at discrete points, and if we count orientation and multiplicitycorrectly, the number of intersection points is independent of our perturbation. We denote thisintersection number by . This gives interesting topological invariants. For example, ifwe embed in via the diagonal embedding, then is the Eulercharacteristic of .

M n X Ydim X + dim Y = n X Y

transversal

(X, Y)IMX X X (X, X) = (X)IXX

X

It turns out that we can describe the intersection numbers in terms of the cohomology ring.Given and , there are corresponding cohomology classes and (called fundamentalclasses), and pairing the with the orientation cycle in the top degree homology spits

X Y [X] [Y][X] [Y]

algebraic topology - What are cohomology rings good for? - Ma... http://math.stackexchange.com/questions/39707/what-are-coho...

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out a number, which is exactly the intersection number of and .X Y

This has also been applied to algebraic geometry with great success. The algebraic geometer inyou may want to look into intersection theory/intersection cohomology.

edited May 19 '11 at 1:24 answered May 17 '11 at 19:40Aaron7,714 2 10 26

You probably want coefficients for your projective space example!2 Grumpy Parsnip May 17 '11 at 19:47

1

@Jim: Yes, you are right. Thanks. Aaron May 18 '11 at 4:42

Just to throw out another example of the use of cup-products, there is an invariant called the, which is the largest such that for some cohomology elements

of degree . The cup length gives a lower bound for the so-called .

cup length k 01 ki 1 Lusternik-Schnirelmanncategory

answered May 17 '11 at 19:53Grumpy Parsnip12.5k 2 16 60

One reason that the cohomology (and homology, and homotopy group) functors are soimportant is that they give a means of testing how similar two spaces are. For instance, the functor suffices to distinguish from , up to homeomorphism (by removing a point); moregenerally, the th homotopy (or homology) group will distinguish from , .

11 2

k k m m > k

Now cohomology, homology, and homotopy groups (and K-theory and other cohomologytheories) are all functors from some category of spaces to some algebraic category. Presumably,the idea is that the algebraic category will be easier to think about -- one can computeexplicitly, at least if the functor applied to nice spaces produces simple results (which oftenhappens).

At the same time, if these are to be a powerful tool at distinguishing spaces, we want thesefunctors to take values not in the simplest category (e.g. the category of sets) but a categorywith more structure: groups, or, better, rings. The cup-product shows that the cohomologyfunctor takes its value in the category of rings.

This means in practice that if you have two spaces with similar looking cohomologies as abeliangroups, you may still be able to tell that they are different if the structures are different. Asan example, you can see that complex projective space and are not homotopy-equivalent, their homology and cohomology groups are all the same. The reason isthat the cup-product is not trivial in the first, while it is trivial in the second.

ring2 S2 S4

even though

So the more algebraic structure you can put on a functor from spaces to sets, the finer thisfunctor will be as a tool to distinguish spaces. Cohomology has this nice ring structure, but infact it has more: it turns out that the cohomology ring (with -coefficients) is a module overthe .

/2Steenrod algebra

answered May 17 '11 at 20:00Akhil Mathew17.4k 1 37 80

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